Suppose I have two simplicial based topological spaces $X_\bullet$ and $Y_\bullet$, and the degeneracy maps of each satisfy the based homotopy extension property (but not necessarily the unbased homotopy extension property. That is, they may not be degreewise wellpointed spaces). Will a simplicial map which is a degreewise weak equivalence induce a weak equivalence on geometric realizations?
This is going to be the same example the one from this previous question: https://mathoverflow.net/a/171423/360
Let $A = \Bbb N$ and $B = \{0,1,1/2,1/3,1/4,\dots\}$, with the map $f: X \to Y$ given as follows: $$ f(n) = \begin{cases} 1/n & n > 0\\ 0 & n = 0 \end{cases} $$ Let $g: X_\bullet \to Y_\bullet$ be the map of simplicial sets obtained by smashing the map $A \to B$ with the simplicial circle $S^1_\bullet$. More specifically, $Y_n$ is an $n$fold wedge $$ B \vee B \vee \cdots \vee B, $$ where degeneracies are an inclusion of all but one wedge factor, and similarly for $X$. On geometric realization, the map $X_\bullet \to Y_\bullet$ is the map from a countable wedge of circles to the Hawaiian earring and is not a weak equivalence. I claim that the degeneracies in both $X$ and $Y$ have the based homotopy extension property.
All of the degeneracies in $X$ have the homotopy extension property full stop because $A$ is wellpointed, so let's examine $Y$. All the degeneracies in $Y$ are of the form $P \to B \vee P$ for $P$ some wedge of copies of $B$ (in particular, $P$ is compact). If $g:B \vee P \to Z$ is a map of based spaces and $H: [0,1] \times P \to Z$ is a based homotopy extending it, then we can simply extend the homotopy to all of $B \vee P$ by letting it be constant on $B$. One can check that this is continuous. (Roughly, this amounts to showing that the map $(B \vee P) \times [0,1] \to B \vee (P \wedge [0,1]_+)$ is a quotient map of spaces and the images of $B$ and $P \times [0,1]$ are closed, but this is automatic since everything in sight is compact Hausdorff.)

$\begingroup$ I think your $A$'s and $B$'s are a bit messed up. The degeneracies in $Y$ are like $P\to B\vee P$, where $P$ is a wedge of copies of $B$. $\endgroup$ – Neil Strickland Oct 14 '15 at 13:55

1$\begingroup$ @NeilStrickland Ach, that was unreadable. I've edited it now. $\endgroup$ – Tyler Lawson Oct 14 '15 at 14:59